Probability

Given that, $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{3}{4}$ and $P\left(C\right)=p$
Probability that the problem can not be solved

$=P\left(\overline{A}\right)\cdot P\left(\overline{B}\right)\cdot P\left(\overline{C}\right)$

$=\left(1-\frac{1}{2}\right)\left(1-\frac{3}{4}\right)(1-p)$

$=\frac{1}{2}\times \frac{1}{4}(1-p)=\frac{1-p}{8}$

$\therefore$ Probability that the problem can be solved

$=1-$Probability that the problem cannot be solved

$\Rightarrow \frac{29}{32}=1-\frac{(1-p)}{8}$

$\Rightarrow 1-p=\frac{3}{4}$

$\therefore p=\frac{1}{4}$

We know that, $P\left(A\mid B\right)=\frac{P(A\cap B)}{P\left(B\right)}$,

$P\left(B\mid A\right)=\frac{P(A\cap B)}{P\left(A\right)}$

Now,

$$\begin{gathered} P\left(A\mid B\right)=P\left(B\mid A\right) \\ \Rightarrow \frac{P(A\cap B)}{P\left(B\right)}=\frac{P(A\cap B)}{P\left(A\right)} \end{gathered}$$

Since $P\left(A\mid B\right)\ne 0$$\Rightarrow P\left(A\cap B\right)\ne 0$.

Therefore,

$$\begin{gathered} P\left(A\cap B\right)·P\left(A\right)=P\left(A\cap B\right)·P\left(B\right) \\ \Rightarrow P\left(A\right)=P\left(B\right) \end{gathered}$$

$P(\overline{E}\mid \overline{F})=\frac{P(\overline{E}\cap \overline{F})}{P\left(\overline{F}\right)}$

$\overline{E}\cap \overline{F}=\overline{E\cup F}$

$P(\overline{E}\cap \overline{F})=P\left(\overline{E\cup F}\right)=1-P(E\cup F)$

$\Rightarrow P(\overline{E}\mid \overline{F})=\frac{1-P(E\cup F)}{P\left(\overline{F}\right)}$

We know that, \(P(S \mid E)=\frac{P(S \cap E)}{P(E)}\)

\(\because S \cap E=E\)

\(\therefore P(S \mid E)=\frac{P(E)}{P(E)}=1 \quad(P(E)>0)\)

$P(A\mid B)=\frac{P(A\cap B)}{P\left(B\right)}$

$P(B\mid A)=\frac{P(A\cap B)}{P\left(A\right)}$

Now,

$$\begin{gathered} P(A\mid B)=P(B\mid A) \\ \frac{P(A\cap B)}{P\left(B\right)}=\frac{P(A\cap B)}{P\left(A\right)} \end{gathered}$$

Since $P(A\mid B)\ne 0$$\Rightarrow P(A\cap B)\ne 0$.

Therefore,

$$\begin{gathered} P(A\cap B)·P\left(A\right)=P(A\cap B)·P\left(B\right) \\ P\left(A\right)=P\left(B\right) \end{gathered}$$

$P\left(\overline{E}\mid \overline{F}\right)=\frac{P\left(\overline{E}\cap \overline{F}\right)}{P\left(\overline{F}\right)}$

Using De morgan's law :

$\overline{E}\cap \overline{F}=\overline{E\cup F}$

$P(\overline{E}\cap \overline{F})=P\left(\overline{E\cup F}\right)=1-P(E\cup F)$

$\Rightarrow P\left(\overline{E}\mid \overline{F}\right)=\frac{1-P(E\cup F)}{P\left(\overline{F}\right)}$

\(P(A \mid B)=P\left(A^{\prime} \mid B\right)\)

\(\frac{P(A \cap B)}{P(B)}=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}\)

\(P(A \cap B)=P\left(A^{\prime} \cap B\right)\)

\(P(A \cap B)=P(B)-P(A \cap B)\)

\(2 P(A \cap B)=P(B)\)

\(\Rightarrow P(A \cap B)=\frac{1}{2} P(B)\)

\(\begin{aligned} & \text { (c) } E _1=\text { denotes selection for } 1^{\text {st }} \text { bag } \\ & E _2=\text { denotes selection for } 2^{\text {st }} \text { bag } \\ & P \left( E _1\right)=\frac{1}{2}, P \left( E _2\right)=\frac{1}{2} \\ & A =\text { selected balls are } 1 \text { red \& } 1 \text { black } \\ & P\left(\frac{A}{E_1}\right)=\frac{{ }^3 C_1 \times C_1}{{ }^6 C_2}=\frac{1}{5} \\ & P\left(\frac{A}{E_2}\right)=\frac{{ }^3 C_1 \times{ }^2 C_1}{(n+5) C_2}=\frac{12}{(n+5)(n+4)} \\ & P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) \times P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \times P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \times P\left(\frac{A}{E_2}\right)} \\ & =\frac{1}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11} \\ & \Rightarrow n=4\end{aligned}\)

\(P(S \mid E)=\frac{P(S \cap E)}{P(E)}\)

\(S \cap E=E\)

\(\Rightarrow P(S \mid E)=\frac{P(E)}{P(E)}=1 \quad(P(E)>0)\)

Given: \(P(A \mid B)=P\left(A^{\prime} \mid B\right)\)

\(\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}\)

\(\Rightarrow P(A \cap B)=P\left(A^{\prime} \cap B\right)\)

\(\Rightarrow P(A \cap B)=P(B)-P(A \cap B)\)

\(\Rightarrow 2 P(A \cap B)=P(B)\)

\(\Rightarrow P(A \cap B)=\frac{1}{2} P(B)\)